An arch of a highway overpass is in the shape of a parabola. The arch spans a distance of 12 meters from one side of the road to the other. The height of the arch is 8 meters at a horizontal distance of 2 meters from each side of the arch. a) Sketch the quadratic function if the vertex of the parabola is on the y-axis and the road is along the x-axis.  b) Use this information to determine the function that models the arch.

### Part (a): Sketching the quadratic function

The parabola representing the arch has the following key features:
1. The **vertex** is at the top of the arch and lies on the \( y \)-axis. Thus, its coordinates are \( (0, 8) \).
2. The parabola is symmetric about the \( y \)-axis.
3. The parabola intersects the \( x \)-axis at points \( (-6, 0) \) and \( (6, 0) \), as the arch spans 12 meters across the road.

With the information provided:
- The road lies along the \( x \)-axis, so the \( x \)-intercepts are at \( (-6, 0) \) and \( (6, 0) \).
- The height of the arch is \( y = 8 \) when \( x = \pm 2 \).

The general form of a parabola is:

\[
y = a x^2 + bx + c
\]

Since the vertex is on the \( y \)-axis, the parabola has no linear term (\( b = 0 \)), making the equation:

\[
y = a x^2 + c
\]

### Part (b): Determining the function

1. Use the vertex \( (0, 8) \):
   At the vertex, \( y = 8 \) when \( x = 0 \). Substituting into the equation \( y = a x^2 + c \):

   \[
   8 = a (0)^2 + c \implies c = 8
   \]

   So, the equation becomes:

   \[
   y = a x^2 + 8
   \]

2. Use the \( x \)-intercepts \( (-6, 0) \) and \( (6, 0) \):
   At these points, \( y = 0 \). Substituting \( x = 6 \) into the equation:

   \[
   0 = a (6)^2 + 8 \implies 0 = 36a + 8 \implies a = -\frac{2}{9}
   \]

   Thus, the equation is:

   \[
   y = -\frac{2}{9}x^2 + 8
   \]

3. Verify using the height of the arch at \( x = \pm 2 \):
   Substituting \( x = 2 \):

   \[
   y = -\frac{2}{9}(2)^2 + 8 = -\frac{2}{9}(4) + 8 = -\frac{8}{9} + 8 = \frac{64}{9}
   \]

   Simplifying, \( y = \frac{64}{9} \approx 7.11 \), which is consistent with the provided height.

### Final Equation:
The function modeling the arch is:

\[
y = -\frac{2}{9}x^2 + 8
\]

### Summary of Steps for Sketching:
1. Draw a symmetric parabola with its vertex at \( (0, 8) \).
2. Label the \( x \)-intercepts at \( (-6, 0) \) and \( (6, 0) \).
3. Verify the shape by ensuring it passes through \( (2, 7.11) \) and \( (-2, 7.11) \).

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### Would you like a graph or additional clarifications?

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### Related Questions:
1. What is the domain and range of the quadratic function representing the arch?
2. How would the equation change if the vertex of the parabola were not at the origin?
3. What is the width of the arch at a height of 6 meters above the road?
4. How does the coefficient \( a \) affect the shape of the parabola?
5. Could this model be used to calculate the load capacity of the arch structure?

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### Tip:
Always verify your parabola by substituting known points to ensure the equation is accurate.

An arch of a highway overpass is in the shape of a parabola. The arch spans a distance of 12 meters from one side of the road to the other. The height of the arch is 8 meters at a horizontal distance of 2 meters from each side of the arch. a) Sketch the quadratic function if the vertex of the parabola is on the y-axis and the road is along the x-axis. b) Use this information to determine the function that models the arch.

Learn how to model a parabolic arch spanning 12 meters with a maximum height of 8 meters using quadratic functions. Step-by-step solution includes determining the equation and verifying with key points. Suitable for high school students studying parabolas and quadratic equations.

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